Question Number 186786 by Mingma last updated on 10/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186787 by Mingma last updated on 10/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186783 by Rupesh123 last updated on 10/Feb/23 Answered by mr W last updated on 10/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186782 by Rupesh123 last updated on 10/Feb/23 Answered by HeferH last updated on 10/Feb/23 $$\:\frac{\mathrm{121}{A}}{\mathrm{60}}−\frac{\mathrm{2}{S}}{\mathrm{3}}\:+\:\frac{\mathrm{121}{A}}{\mathrm{40}}−{S}\:=\:\frac{\mathrm{17}{A}}{\mathrm{8}} \\ $$$$\:\frac{\mathrm{121}{A}}{\mathrm{24}}−\frac{\mathrm{5}{S}}{\mathrm{3}}\:=\:\frac{\mathrm{17}{A}}{\mathrm{8}} \\ $$$$\:\frac{\mathrm{121}{A}}{\mathrm{24}}−\frac{\mathrm{17}\centerdot\mathrm{3}{A}}{\mathrm{8}\centerdot\mathrm{3}}=\:\frac{\mathrm{5}{S}}{\mathrm{3}} \\ $$$$\:\frac{\mathrm{70}{A}}{\mathrm{24}}=\:\frac{\mathrm{5}{S}}{\mathrm{3}} \\ $$$$\:\frac{\mathrm{70}{A}\centerdot\mathrm{3}}{\mathrm{24}\centerdot\mathrm{5}}\:=\:{S}…
Question Number 186764 by HeferH last updated on 09/Feb/23 Commented by HeferH last updated on 09/Feb/23 $${find}\:{x}\: \\ $$ Commented by Frix last updated on…
Question Number 186762 by ajfour last updated on 09/Feb/23 Commented by ajfour last updated on 09/Feb/23 $${What}\:{minimum}\:{length}\:{to}\:{walk}\: \\ $$$${to}\:{start}\:{from}\:{home}\:{with}\:{empty} \\ $$$${bucket}\:{go}\:{get}\:{water}\:{from}\:{well}\:{n} \\ $$$${water}\:{the}\:{tree}. \\ $$…
Question Number 186697 by Mingma last updated on 08/Feb/23 Answered by mr W last updated on 08/Feb/23 $${DB}=\frac{\mathrm{8}\:\mathrm{sin}\:\mathrm{80}°}{\mathrm{sin}\:\mathrm{30}°} \\ $$$${x}=\frac{{DB}×\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:\:=\frac{\mathrm{8}\:\mathrm{sin}\:\mathrm{80}°\:\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:\:=\frac{\mathrm{8}\:\mathrm{cos}\:\mathrm{10}°\:\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{20}°} \\…
Question Number 186698 by Mingma last updated on 08/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186667 by cherokeesay last updated on 08/Feb/23 Answered by som(math1967) last updated on 08/Feb/23 $${cos}\mathrm{30}=\frac{{QR}^{\mathrm{2}} +{QT}^{\mathrm{2}} −{RT}^{\mathrm{2}} }{\mathrm{2}×{QR}×{QT}} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{16}+\mathrm{16}−{RT}^{\mathrm{2}} }{\mathrm{2}×\mathrm{16}} \\ $$$$\Rightarrow{RT}=\sqrt{\mathrm{32}−\mathrm{16}\sqrt{\mathrm{3}}}=\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}…
Question Number 121134 by behi83417@gmail.com last updated on 05/Nov/20 Answered by ajfour last updated on 05/Nov/20 $${must}\:{be}\:{point}\:{locus}. \\ $$ Commented by behi83417@gmail.com last updated on…